The Circle Exercise: 11.1 Class 11 Basic Mathematics Solution [NEB UPDATED]

Exercise 11.1

1) Find the equations of the circles with:

a) Centre at (4,-1) and through the origin

Solution:

Given centre = (4, 1), radius = $\sqrt {{{\left( {4 - 0} \right)}^2} + {{\left( { - 1 - 0} \right)}^2}} $ = $\sqrt {16 + 1} $ = $\sqrt {17} $.

Equation of circle is (x – 4)² + (y  + 1)² = $\left( {\sqrt {17} } \right)$².

Or, x² – 8x + 16 – y² + 2y + 1 = 17

So, x² + y² – 8x + 2y = 0 is the required equation

b) Two of the diameters are x+y=6 and x+2y=8 and radius 10.

Solution:

Solving x + y = 6 and x + 2y = 8, we get,

X = 4, y = 2.

Centre = (4, 2).

Equation of circle is (x – 4)² + (y – 2)² = 10²

Or, x² – 8x + 16 – y² – 4y + 25 – 100 = 0

So, x² + y² – 8x – 4y – 80 = 0 is the required equation

c) Center at (-1, 5) and through the point of intersection of the lines 2x - y = 5 and 3x + y = 10

Solution:

Solving 2x – y = 5 and 3x + y = 10, we get,

X = 3, y = 1.

The circle through the point (3, 1).

Radius = $\sqrt {{{\left( {3 + 1} \right)}^2} + {{\left( {1 - 5} \right)}^2}} $ = $\sqrt {32} $.

Equation of circle at (–1.5) with radius $\sqrt {32} $is

(x + 1)² + (y – 5)² = ($\sqrt {32} $)²

Or, x² + 2x + 1 + y² – 10y + 25 = 32

So, x² + y² + 2x – 10y = 6  is the required equation of the circle.

d) With (0, 0) and (4, 7) as the ends of a diameter

Solution:

Equation of circle in diameter form is:

Or, (x – 0)(x – 4) + (y – 0)(y – 7) = 0

Or, x² – 4x + y² – 7y = 0

So, x² + y² – 4x – 7y = 0 is the required equation of the circle.

e) Concentric with the circle x² + y ² + 8x - 6y + 1 = 0 and radius 3

Solution:

Comparing the equation with x² + y² + 2gx + 2fy + c =0

g = 4, f = -3, c = 1

Centre = (-g,-f) = (-4,3)
Equation of circle centre at (-4, 3) with radius 3 is

Or, (x + 4)² + (y – 3)² = 3²

Or, x² + y² + 8x – 6y + 16 = 0, is the required eqation.

f) Concentric with the circle x² + y ²- 8x + 12y + 15 = 0 and passing through (5, 4)

Solution:

Comparing the equation with x² + y² + 2gx + 2fy + c =0

g = –4, f= 6, c = 15.

Centre = (–g, –f) = (4, –6)

Radius = $\sqrt {{{\left( {5 - 4} \right)}^2} + {{\left( {4 + 6} \right)}^2}} $ = $\sqrt {1 + 100} $ = $\sqrt {101} $.

Equation of circle centre at (4, –6) with radius = $\sqrt {101} $ is

Or, (x – 4)² + (y + 6)² = ${\left( {\sqrt {101} } \right)^2}$.

Or, x² – 8x + 16 + y² + 12y + 36 = 101

Or, x² + y² – 8x + 12y= 49 is the required equation of the circle.

g) Passing through the origin and making intercepts equal to 3 and 4 from the positive and y-axis respectively
Solution:

The circle through the points (0, 0), (3, 0) and (0, 4).

Let the equation of circle be x² + y² + 2gx + 2fy + c = 0.

Then, c = 0.

Or, 9 + 6g + 0 = 0 → g = $ - \frac{3}{2}$.

And 0 + 16 + 8f = 0 → f = –2.

Therefore, x² + y² + 2$\left( { - \frac{3}{2}} \right)$x + 2(–2).y + 0 = 0

So, x² + y² – 3x – 4y = 0 is the required equation.

2) Find the equation of the circle

a) With centre at (3, 4) and touching the x-axis

Solution:

If the circle touching x–axis then radius = 4.

Where, centre = (3, 4)

Equation of circle centre (3, 4) and radius = 4 is:

Or, (x – 3)² + (y –  4)² = 16

Or, x² – 6x + 9 + y² – 8y + 16 = 16.

So, x² + y² – 6x – 8y + 9 = 0 is the required equation of the circle.

b) With centre at (a, b) and touching the y-axis

Solution:

If the circle touching y–axis then radius = a.

Where, centre = (a, b)

Equation of circle centre (a, b) and radius = a is:

Or, (x – a)² + (y –  b)² = a².

Or, x² – 2ax + a² + y² – 2by + b² = a².

So, x² + y² – 2ax – 2by + b² = 0 is the required equation of the circle.

c) With centre at (4, 5) and touching the line 3x - 4y + 18 = 0

Solution:

Radius = $ \pm \frac{{3.4 - 4.5 + 18}}{{\sqrt {9 + 16} }}$ = $ \pm \frac{{10}}{5}$ = ±2

So, radius = 2.

Equation of circle centre (4, 5) with radius 2 is:

Or, (x – 4)² + (y – 5)² = 4.

Or, x² – 8x + 16 + y² – 10y+ 25 = 4

So, x² + y² – 8x – 10y + 37 = 0 is the required equation of the circle.

d) With centre on the line 3x - 5y = 4 and touching both axes of coordinates

Solution:

Let (h, k) be the centre of the circle. If the circle touching both axes then h = k = Radius.

Also, 3h – 5h = 4

Or, –2h = 4

So, h = –2

Now the equation of the circle centre at (–2, –2) and radius 2 isL

Or, (x + 2)² + (y + 2)² = 4

Or, x² + 4x + 4 + y² + 4y + 4 = 4

So, x² + y² + 4x + 4y + 4 = 0 is the required equation

e) Which touches each axis and passes through the point (2, 1)

Solution:

Let (h, k) be the centre of the circle. IF the circle ouches each axis then radius = h= k.

So, $(\sqrt {{{\left( {{\rm{h}} - 2} \right)}^2} + {{\left( {{\rm{h}} - 1} \right)}^2}} $ = h

Or, h² – 4h + 4 + h² – 2h + 1 = h².

Or, h² – 6h + 5 = 0.

Or, (h – 5)(h – 1) = 0

So, h = 5 or, h = 1.

Now the equation of circle centre at(5, 5) and radius 5 is:

Or, (x – 5)² + (y – 5)² = 5².

Or, x² – 10x + 25 + y² – 10y + 25 = 25.

So, x² + y² – 10x – 10y + 25 = 0.

Again, the equation of the circle centre at (1, 1) and radius = 1 is:

(x – 1)² + (y – 1)² = 1²

Or, x² – 2x + 1 + y² – 2y + 1 = 1

So, x² + y² – 2x – 2y + 1 =0

So, x² + y² – 10x – 10y + 25 = 0 and x² + y² – 2x – 2y + 1 = 0 are the required equation of circle.

f) Which touches the coordinate axes at (a, 0) and (0, a)

Solution:

If the circle touches the co–ordinate at (a, 0) and (0, a) then centre = (a, a) and radius = a.

Now, the equation of circle centre at(a, a) with radius = a is

Or, (x – a)² + (y – a)² = a²

Or, x² – 2ax + a² + y² – 2ay + a² = a².

So, x² + y² – 2ax – 2ay + a² = 0 is the required equation if circle.

g) With radius 4, center on the line 13x + 4y = 32 and touching the line 3x + 4y = 12

Solution:

Let (h, k) be the centre of the circle,

Then, 13h + 4k = 32….(i)

The length of the perpendicular from (h, k) to the tangent line 3x + 4y – 12 = 0 is 4.

So, 4 = $ \pm \left[ {\frac{{3{\rm{h}} + 4{\rm{k}} - 12}}{{\sqrt {9 + 16} }}} \right]$ = $ \pm \left[ {\frac{{3{\rm{h}} + 4{\rm{k}} - 12}}{5}} \right]$

Taking +ve sign, 3h + 4k = 32 …(ii)

Taking –ve sign, 3h + 4k = –8 ….(iii)

Sloving (i) and (ii) we get,

h = 0, k =8.

The equation of circle at (0, 8) and radius 4 is:

Or, (x – 0)² + (y – 8)² = 16

Or, x² + y² – 16y + 64 = 16

Or, x² + y² – 16y + 48 = 0

Again solving, (i) and (iii) we get,

h = 4, k = –5.

The equation of circle centre at (4, –5) and radius with 4 is:

Or, (x – 4)² + (y + 5)² = 16

Or, x² – 8x + 16 + y² + 10y + 25 = 16

Or, x² + y² – 8x + 10y + 25 = 0

So, x² + y² – 16y + 48 = 0 and x² + y² – 8x + 10y + 25 = 0 are the required equations.

3. Find the center and the radius of the following circles

a) x² + y ² - 12x - 4y = 9

Solution:
Compare the given equation with

x² + y² + 2gx + 2fy + c = 0

g = –6 , f = –2, c = –9.

So, Centre = (–g, –f) = (6, 2)

and, Radius = $\sqrt {{{\rm{g}}^2} + {{\rm{f}}^2}} $ = $\sqrt {36 + 4 + 9} $ = 7.
b) x² + y ² - 3x + 2y - 3/4 = 0

Solution:

Compare the given equation with

x² + y² + 2gx + 2fy + c = 0

g = $ - \frac{3}{2}$, f = 1, c = $ - \frac{3}{4}$.

So, Centre = (–g, –f) = ($\frac{3}{2}$, –1)

and, Radius = $\sqrt {{{\rm{g}}^2} + {{\rm{f}}^2} - {\rm{c}}} $ = $\sqrt {\frac{9}{4} + 1 + \frac{3}{4}} $ = 2.

4. Find the equation of the circle

a) passing through the points (0, 0). (a, 0) and (0, b)

Solution:
Let x² + y² + 2gx + 2fx + c = 0 …(i) be the equation of the circle.

If (i) passes through (0, 0), then c = 0.

If (i) passes through (a, 0), then a² + 2ag = 0 à g = $ - \frac{{\rm{a}}}{2}$.

If (i) passes through (0, b) then b² + 2bf = 0 à f = $ - \frac{{\rm{b}}}{2}$.

Putting the values of g, f, c in equation (i), we get,

Or, x² + y² + 2$\left( { - \frac{{\rm{a}}}{2}} \right)$x + 2 $\left( { - \frac{{\rm{b}}}{2}} \right)$y + 0 = 0.

So, x² + y² – ax – by = 0 is the required equation of circle.
b) Through the points (1, 2), (3, l) and (-3,-1)

Solution:

Let x² + y² + 2gx + 2fy + c = 0 …(i) be the equation of the circle.

If (i) passes through (1, 2) ,(3, 1) and (–3, –1) then,

2g + 4f + c + 5 = 0 …(ii)

6g + 2f + c + 10 = 0 ….(iii)

–6g – 2f + c + 10 = 0 …(iv)

Solving (ii), (iii) and (iv), we get,

g = $ - \frac{1}{2}$, f = $\frac{3}{2}$, c = –10

Putting these values in equation (i) we get,

Or, x² + y² + 2.$\left( { - \frac{1}{2}} \right)$x + 2. $\frac{3}{2}$y – 10 = 0

So, x² + y² – x + 3y – 10 is the required equation of the circle.

5. Show that the points (3, 3), (6, 4), (7, 1) and (4, 0) are concyclic. Find the center and the radius of the circle.

Solution:

Let x² + y² + 2gx + 2fy + c = 0….(i) be the equation of the circle.

If it passes through (3, 3), (6, 4) and (7, 1) then,

Or, 6g + 6f + c + 18 = 0 ….(ii)

Or, 12g + 8f + c + 52 = 0 …(iii)

Or, 14g + 2f + c + 50 = 0 …..(iv)

Solving (ii), (iii) and (iv).

g = –5 , f = –2 , c = 24.

Putting these values in equation (i), we get,

Or, x² + y² + 2(–5)x + 2(–2)y + 24 = 0

So, x² + y² – 10x – 4y + 24 = 0 is the equation of the circle.

Putting the fourth point (4, 0) in the equation of circle.

Or, 16 + 0 – 40 – 0 + 24 = 0

Or, 0 = 0

So, (4, 0) lies on the circle.

Hence, the given four points are concyclic.

Centre = (–g, –f) = (5, 2)

And radius = $\sqrt {{{\rm{g}}^2} + {{\rm{f}}^2} - {\rm{c}}} $ = $\sqrt {25 + 4 - 24} $ = $\sqrt 5 $.